I'm trying to find why the linear function f(-x)=x√x^2+2 is negative when it should be neither. why are √'s not negatives?!?!?! Is it because √x can't cross over the y axis because √x can't be negative? Then why is it that the x on the outside of the function is negative and the whole thing is counted as a neither?!

First of all your function is not a linear function, (it is not a straight line)

Secondly, is it the way you typed it or is it
f(-x) = x√(x^2 + 2)

by the way,
f(x) = -x√(x^2+2)

for negative x's the graph rises in quadrant II
for positive x's the graph drops in quadrant IV
it crosses the x-axis at (0,0)

Wolfram shows this:
http://www.wolframalpha.com/input/?i=-x√%28x%5E2%2B2%29

To understand why the function f(-x) = x√(x^2 + 2) is negative when it should be neither negative nor positive (neither), let's break it down step by step.

First, let's start with the square root (√). It is true that the square root of a positive number is always positive. Therefore, the term √x^2 would always be positive for any value of x, as x^2 is non-negative.

Now, let's move on to the negative sign outside the function. When you substitute -x into the function f(-x), it becomes f(-x) = (-x)√((-x)^2 + 2).

Since we are taking the square root of a positive value, the term inside the square root, (-x)^2 + 2, will also be positive.

Now, when we multiply the positive term inside the square root by the negative sign (-x), the overall result becomes negative. This is why f(-x) is negative when x is substituted into the function.

However, it is important to note that the negative value of f(-x) does not necessarily mean that the function is negative as a whole or that it cannot cross the y-axis. The negativity is only related to the specific value of x you have chosen.

To determine whether the function f(x) = x√(x^2 + 2) is negative, positive, or neither for all values of x, you would need to analyze the behavior of the function as x approaches positive or negative infinity by using calculus techniques or graphical analysis.